The angle between the line `(x+1)/(2)=(y)/(3)=(z-3)/(-6)` and the plane `10x + 2y + 11z=8` is

Options

A. `sin^(-1)((4)/(21))`
B. `sin^(-1)((4)/(7))`
C. `sin^(-1)((8)/(21))`
D. `sin^(-1)((8)/(7))`

Answer

To find the angle between the line given by the equation \((x+1)/(2)=(y)/(3)=(z-3)/(-6)\) and the plane given by the equation \(10x + 2y + 11z = 8\), we can follow these steps:

### Step 1: Identify the direction ratios of the line
The line is represented in symmetric form. From the equation \((x+1)/(2)=(y)/(3)=(z-3)/(-6)\), we can extract the direction ratios of the line:
- The direction ratios are \(2\), \(3\), and \(-6\).

Thus, we can represent the direction vector of the line as:
\[ \mathbf{A} = \langle 2, 3, -6 \rangle \]

### Step 2: Identify the normal vector of the plane
The equation of the plane is given as \(10x + 2y + 11z = 8\). The coefficients of \(x\), \(y\), and \(z\) give us the normal vector of the plane:
- The normal vector is \(\mathbf{B} = \langle 10, 2, 11 \rangle\).

### Step 3: Calculate the dot product of the vectors
The dot product \(\mathbf{A} \cdot \mathbf{B}\) is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (2)(10) + (3)(2) + (-6)(11) \]
Calculating each term:
- \(2 \cdot 10 = 20\)
- \(3 \cdot 2 = 6\)
- \(-6 \cdot 11 = -66\)

Now, summing these results:
\[ \mathbf{A} \cdot \mathbf{B} = 20 + 6 - 66 = -40 \]

### Step 4: Calculate the magnitudes of the vectors
Now, we need to find the magnitudes of both vectors:
\[ |\mathbf{A}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]
\[ |\mathbf{B}| = \sqrt{10^2 + 2^2 + 11^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15 \]

### Step 5: Use the formula for the sine of the angle
The sine of the angle \(\theta\) between the line and the plane is given by:
\[ \sin \theta = \frac{|\mathbf{A} \cdot \mathbf{B}|}{|\mathbf{A}| \cdot |\mathbf{B}|} \]
Substituting the values we calculated:
\[ \sin \theta = \frac{|-40|}{7 \cdot 15} = \frac{40}{105} = \frac{8}{21} \]

### Step 6: Calculate the angle \(\theta\)
To find the angle \(\theta\), we take the inverse sine:
\[ \theta = \sin^{-1}\left(\frac{8}{21}\right) \]

### Final Answer
Thus, the angle between the line and the plane is:
\[ \theta = \sin^{-1}\left(\frac{8}{21}\right) \]